Solubility approximation

We have calculated the solubility by saying that if there is 2 × 10^-5 mol/100 g of water then there must be ten times as much in 1000 g of water.

2 × 10^-5 × 10 = 2 × 10^-4 mol/1000 g of water

1000 g of water has a volume of 1000 cm³ or 1 dm³. However, this is not strictly the same as concentration in mol dm^-3. We are saying that 1000 cm³ of water will dissolve 2 × 10^-4 moles of calcium hydroxide, but a solubility of 2 × 10^-4 mol dm^-3 means that 1000 cm³ of solution contains 2 × 10^-4 mol of calcium hydroxide. It's rather a subtle point, particularly where the solubility is so low as it is in this case. The tiny amount of calcium hydroxide dissolved in 1000 cm³ of water will have very little effect on the overall volume of the solution. So, in this case, it is a very good approximation.